Tag: color int determines IDT width share shadow print void
The
problem description places n adjacent rectangles on the horizontal axis, each rectangle having a width of 1, and the height of the rectangle I (1≤i≤n) is hi. These n rectangles form a histogram. For example, the height of the six rectangles is 3, 1, 6, 5, 2, 3, respectively.
Find the rectangle with the largest area in the given histogram, and its edges are parallel to the axis. For the example given above, the maximum rectangle as shown in the shaded portion, the area is 10.
Input format the first line contains an integer n, which is the number of rectangles (1≤n≤1000).
The second line contains n integers h1, h2, ..., HN, and the adjacent numbers are separated by a space. (1≤hi≤10000). Hi is the height of the first rectangle. The output format outputs a row that contains an integer that is the area of the largest rectangle within a given histogram. Sample Input 6
3 1 6 5 2 3 Sample Output
The idea of solving problems: Enumerate all the rectangles that can form the requirements, select the maximum output. It is important to note that the height of the rectangle is determined by the smallest integer hi, so you need to find the minimum hi. Code See below:
1#include <stdio.h>2 #defineNmax 10103 4 intAeraintA[],intn);5 intMainvoid)6 {7 inti, N;8 intA[nmax];9scanf"%d",&n);Ten for(i=0; I < n; i + +) One { Ascanf"%d",&a[i]); - } -printf"%d", Aera (a,n)); the return 0; - } - intAeraintA[],intlen) - { + intI, low,temp, Max_aera =-1; - for(i =0; I <len; i++) + { Alow = A[i];//set minimum integer to low at for(intj = i; J <len; J + +) -{//take position I as the starting point and position J as the end point to find the rectangle that meets the requirements. The beginning of each round is unchanged, the end point +1 - if(A[j] <Low ) -Low =A[j]; -Temp = (J-i +1) * LOW;//calculation formula of rectangular area - if(Temp >Max_aera) inMax_aera =temp; - } to } + returnMax_aera; -}
CCF2013123 the largest rectangle (C language version)